A066983 -id:A066983 - cp's OEIS frontend

A121364 Convolution of A066983 with the double Fibonacci sequence A103609.

0, 0, 1, 2, 3, 6, 10, 18, 29, 50, 81, 136, 220, 364, 589, 966, 1563, 2550, 4126, 6710, 10857, 17622, 28513, 46224, 74792, 121160, 196041, 317434, 513619, 831430, 1345282, 2177322, 3522981, 5701290, 9224881, 14927768, 24153636, 39083988, 63239221, 102327390

Offset: 1

Graeme McRae, Jul 23 2006

The convolution of 1,0,1,1,1,3,3,7,9,17,25,... (A066983 with 1,0 added to the front) with "A double Fibonacci sequence" (A103609) is the Fibonacci sequence (A000045), with an extra initial 0.

			a(7)=10 because F(7)=13 and D(8)=3 and a(7)=F(7)-D(8).

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,0,-1,-1).

Cf. A000045, A066983, A103609.

A121364:= func< n | Fibonacci(n) - Fibonacci(Floor((n+1)/2)) >;
[A121364(n): n in [1..70]]; // G. C. Greubel, Oct 23 2024

LinearRecurrence[{1,2,-1,0,-1,-1},{0, 0, 1, 2, 3, 6},40] (* James C. McMahon, Oct 17 2024 *)

concat([0,0], Vec(-x^3*(x^2-x-1)/((x^2+x-1)*(x^4+x^2-1)) + O(x^100))) \\ Colin Barker, Oct 13 2014

def A121364(n): return fibonacci(n) - fibonacci((n+1)//2)
[A121364(n) for n in range(1,71)] # G. C. Greubel, Oct 23 2024

a(n) = F(n) - D(n+1), where F is the Fibonacci sequence (A000045) and D is "A double Fibonacci sequence" (A103609).

G.f.: x^3*(1+x-x^2) / ((1-x-x^2)*(1-x^2-x^4)). - Colin Barker, Oct 13 2014

More terms from Colin Barker, Oct 13 2014

A153864 Triangle read by rows, A000012 * A153860 * (A066983 * 0^(n-k)).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 2, 2, 3, 2, 2, 2, 6, 3, 1, 2, 2, 6, 6, 7, 2, 2, 2, 6, 6, 14, 9, 1, 2, 2, 6, 6, 14, 18, 17, 2, 2, 2, 6, 6, 14, 18, 34, 25, 1, 2, 2, 6, 6, 14, 18, 34, 50, 43, 2, 2, 2, 6, 6, 14, 18, 34, 50, 86, 67

Offset: 0

Gary W. Adamson, Jan 03 2009

Row sums = A066629: (1, 2, 5, 8, 15, 24, 41, 66, 109,...).

Right border = A066983: (1, 1, 1, 3, 3, 7, 9, 17,...).

			First few rows of the triangle =
1;
1, 1;
2, 2, 1;
1, 2, 2, 3;
2, 2, 2, 6, 3;
1, 2, 2, 6, 6, 7;
2, 2, 2, 6, 6, 14, 9;
1, 2, 2, 6, 6, 14, 18, 17;
2, 2, 2, 6, 6, 14, 18, 34, 25;
1, 2, 2, 6, 6, 14, 18, 34, 50, 43;
...

Cf. A153860, A066983, A066629

Triangle read by rows, A000012 * A153860 * (A066983 * 0^(n-k))
Given triangle A000012 * A153860 = partial sums of A153860 starting from the top.
(A066983 * 0^n-k) = an infinite lower triangular matrix with A066983 as the
main diagonal: (1, 1, 1, 3, 3, 7, 9, 17, 25,...) and the rest zeros.

A111091 Successive generations of a Kolakoski(3,1) rule starting with 1 (see A066983).

Original entry on oeis.org

1, 3, 111, 313, 1113111, 313111313, 11131113131113111, 3131113131113111313111313, 1113111313111311131311131311131113131113111

Offset: 1

Benoit Cloitre, Oct 12 2005

nonn

Terms are palindromic. If b_3(n) denotes the number of 3's in a(n) then b(n) satisfies the recursion: b_3(1)=0, b_3(2)=1 and b_3(n) = b_3(n-1) + b_3(n-2) + (-1)^n so that b_3(2n)=A055588(n) and b_3(2n+1)=A027941(n). If b_1(n) denotes the number of 1's: b_1(1)=1, b_1(2)=0 and b_1(n) = b_1(n-1) + b_1(n-2) - 2*(-1)^n so that b_1(2n)=A004146(n) and b_1(2n+1) = A000032(n-2) - 2.

			1 --> 3 --> 111 --> 313 --> 1113111 --> 313111313

Cf. A111081.

As n grows, a(2n-1) converges toward A095345 (read as a word) and a(2n) converges toward A095346.

A052952 a(n) = Fibonacci(n+2) - (1-(-1)^n)/2.

Original entry on oeis.org

1, 1, 3, 4, 8, 12, 21, 33, 55, 88, 144, 232, 377, 609, 987, 1596, 2584, 4180, 6765, 10945, 17711, 28656, 46368, 75024, 121393, 196417, 317811, 514228, 832040, 1346268, 2178309, 3524577, 5702887, 9227464, 14930352, 24157816, 39088169, 63245985, 102334155

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Equals row sums of triangle A173284. - Gary W. Adamson, Feb 14 2010
The Kn21 sums (see A180662 for definition) of the 'Races with Ties' triangle A035317 produce this sequence. - Johannes W. Meijer, Jul 20 2011
a(n-1), for n >= 1, gives the number of compositions of n with relative prime parts, and parts not exceeding 2. See the row sums of triangle A030528 where for even n the leading 1 is missing. - Wolfdieter Lang, Jul 27 2023

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 8*x^4 + 12*x^5 + 21*x^6 + 33*x^7 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1023
K. Kuhapatanakul, On the Sums of Reciprocal Generalized Fibonacci Numbers, J. Int. Seq. 16 (2013) #13.7.1, eq (1).
Steven Linton, James Propp, Tom Roby, and Julian West, Equivalence Classes of Permutations under Various Relations Generated by Constrained Transpositions, Journal of Integer Sequences, Vol. 15 (2012), #12.9.1.
H. Ohtsuka and S. Nakamura, On the sum of reciprocal sums of Fibonacci numbers, Fibonacci Quart. 46/47 (2008/2009), 153-159.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).

Cf. A062114, A173284, A059841, A014217, A180662, A035317.
Partial sums of A008346, first differences of A129696.
Cf. also A000032, A000045, A030528.
Cf. A001595, A054450, A066983, A074331, A168309.

List([0..40], n-> Fibonacci(n+2) -(1-(-1)^n)/2); # G. C. Greubel, Jul 10 2019

a052952 n = a052952_list !! n
a052952_list = 1 : 1 : zipWith (+)
   a059841_list (zipWith (+) a052952_list $ tail a052952_list)
-- Reinhard Zumkeller, Jan 06 2012

[Fibonacci(n+2)-(1-(-1)^n)/2: n in [0..40]]; // Vincenzo Librandi, Dec 02 2016

A052952 :=proc(n)
    option remember;
    local t1;
    if n <= 1 then
        return 1 ;
    fi:
    if n mod 2 = 1 then
        t1:=0
    else
        t1:=1;
    fi:
    procname(n-1)+procname(n-2)+t1;
end proc;
seq(A052952(n), n=0..40) ; # N. J. A. Sloane, May 25 2008

Table[Fibonacci[n+2] -(1-(-1)^n)/2, {n, 0, 40}] (* Vincenzo Librandi, Dec 02 2016 *)
Sum[(-1)^k*Fibonacci[Range[2,41], 1-k], {k,0,1}] (* G. C. Greubel, Oct 21 2019 *)
CoefficientList[Series[1/((1-x-x^2)*(1-x^2)),{x,0,40}],x] (* Harvey P. Dale, Sep 12 2020 *)

PARI
```
{a(n) = fibonacci(n+2) - n%2};
```

[fibonacci(n+2) -(1-(-1)^n)/2 for n in (0..40)] # G. C. Greubel, Jul 10 2019

G.f.: 1/((1-x-x^2)*(1-x^2)).
a(n) = A074331(n+1).
a(n) = A054450(n+1, 1) (second column of triangle).
a(n) = 2*a(n-2) + a(n-3) + 1, with a(0)=1, a(1)=1, a(2)=3.
a(n) = Sum_{alpha=RootOf(-1+z+z^2)} (3+alpha)*alpha^(-1-n)/3 - Sum_{beta=RootOf(-1+z^2)} beta^(-1-n)/2.
a(2*k) = Sum_{j=0..k} F(2*j+1) = F(2*(k+1)) for k >= 0; a(2*k-1) = Sum_{j=0..k} F(2*j) = F(2*k+1)-1 for k >= 1 (F = A000045, Fibonacci numbers).
a(n) = a(n-1) + a(n-2) + (1+(-1)^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+1, k). - Paul Barry, Oct 23 2004
a(n) = floor(phi^(n+2) / sqrt(5)), where phi is the golden ratio: phi = (1+sqrt(5))/2. - Reinhard Zumkeller, Apr 19 2005
a(n) = Fibonacci(n+1) + a(n-2) with n>1, a(0)=a(1)=1. - Zerinvary Lajos, Mar 17 2008
a(n) = floor(Fibonacci(n+3)^2/Fibonacci(n+4)). - Gary Detlefs, Nov 29 2010
a(n) = (A001595(n+3) - A066983(n+4))/2. - Gary Detlefs, Dec 19 2010
a(4*n) = F(4*n+2); a(4*n+1) = F(4*n+3) - 1; a(4*n+2) = F(4*n+4); a(4*n+3) = F(4*n+5) - 1. - Johannes W. Meijer, Jul 20 2011
a(n+1) = a(n) + a(n-1) + A059841(n+1). - Reinhard Zumkeller, Jan 06 2012
a(n) = floor(|F((1+i)*(n+2))|), n >= 0, with the complex Fibonacci function F: C -> C, z -> F(z) with F(z) := (exp(log(phi)*z) - exp(i*Pi*z)*exp(-log(phi)*z))/(2*phi-1) with the modulus |z|, the imaginary unit i and the golden section phi:=(1+sqrt(5))/2. A Conjecture: For F(z) see, e.g., the T. Koshy reference. ch. 45, p. 523, where F is called f, given in A000045. - Wolfdieter Lang, Jul 24 2012
5*a(n) = (L(n+3)-1)*(L(n+4)+3) -14 -Sum_{k=0..n} L(k+1)*L(k+5) = (L(n+3)-1)*(L(n+4)+3) -L(2*n+7) +A168309(n), where L=A000032. - J. M. Bergot, Jun 13 2014
a(n) = floor(phi*Fibonacci(n+1)), where phi is the golden section. - Michel Dekking, Dec 02 2016
a(n) = -(-1)^n * a(-4-n) for all n in Z. - Michael Somos, Dec 03 2016
a(n) = Sum_{k=0..n} Sum_{i=0..n} C(n-k-1,k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = floor(1/(Sum_{k>=n+4} 1/Fibonacci(k))) [Ohtsuka and Nakamura]. - Michel Marcus, Aug 09 2018
a(n) = floor(abs(chebyshevU(n/2, 3/2))). - Federico Provvedi, Feb 23 2022
E.g.f.: exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5 - sinh(x). - Stefano Spezia, Mar 09 2024

Additional formulas and more terms from Wolfdieter Lang, May 02 2000

Better description from Olivier Gérard, Jun 05 2001

A008346 a(n) = Fibonacci(n) + (-1)^n.

Original entry on oeis.org

1, 0, 2, 1, 4, 4, 9, 12, 22, 33, 56, 88, 145, 232, 378, 609, 988, 1596, 2585, 4180, 6766, 10945, 17712, 28656, 46369, 75024, 121394, 196417, 317812, 514228, 832041, 1346268, 2178310, 3524577, 5702888, 9227464, 14930353, 24157816, 39088170, 63245985, 102334156

Offset: 0

N. J. A. Sloane

Diagonal sums of A059260. - Paul Barry, Oct 25 2004
The absolute value of the Euler characteristic of the Boolean complex of the Coxeter group A_n. - Bridget Tenner, Jun 04 2008
a(n) is the number of compositions (ordered partitions) of n into two sorts of 2's and one sort of 3's.  Example: the a(5)=4 compositions of 5 are 2+3, 2'+3, 3+2 and 3+2'. - Bob Selcoe, Jun 21 2013
Let r = 0.70980344286129... denote the rabbit constant A014565. The sequence 2^a(n) gives the simple continued fraction expansion of the constant r/2 = 0.35490172143064565732 ... = 1/(2^1 + 1/(2^0 + 1/(2^2 + 1/(2^1 + 1/(2^4 + 1/(2^4 + 1/(2^9 + 1/(2^12 + ... )))))))). Cf. A099925. - Peter Bala, Nov 06 2013
a(n) is the top left entry of the n-th power of the 3 X 3 matrix [0, 1, 1; 1, 0, 1; 1, 0, 0] or of the 3 X 3 matrix [0, 1, 1; 1, 0, 0; 1, 1, 0]. - R. J. Mathar, Feb 03 2014
a(n) is the number of growing self-avoiding walks with n+3 edges on the grid graph of integer points (x,y) with x >= 0 and y in {0, 1} and with a trapped endpoint. - Jay Pantone, Jul 26 2024

			The Boolean complex of Coxeter group A_4 is homotopy equivalent to the wedge of 2 spheres S^3, which has Euler characteristic 1 - 2 = -1.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
G. Bilgici, Generalized order-k Pell-Padovan-like numbers by matrix methods, Pure and Applied Mathematics Journal, 2013; 2(6): 174-178.
Tomislav Došlić, Mate Puljiz, Stjepan Šebek, and Josip Žubrinić, Predators and altruists arriving on jammed Riviera, arXiv:2401.01225 [math.CO], 2024. See p. 14.
N. Gogin and A. Mylläri, Padovan-like sequences and Bell polynomials, Proceedings of Applications of Computer Algebra ACA, 2013.
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See pp. 4, 13.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 445
Jay Pantone, A. R. Klotz, and E. Sullivan, Exactly-solvable self-trapping lattice walks. II. Lattices of arbitrary height., arXiv:2407.18205 [math.CO], 2024.
K. Ragnarsson and B. E. Tenner, Homotopy type of the Boolean complex of a Coxeter system, arXiv:0806.0906 [math.CO], 2008-2009.
Index entries for linear recurrences with constant coefficients, signature (0,2,1).

Cf. A007492, A066983, A078024, A119282, A014565, A099925, A098600, A374297.

List([0..50], n-> Fibonacci(n) + (-1)^n); # G. C. Greubel, Jul 13 2019

[Fibonacci(n) + (-1)^n: n in [0..50]]; // Vincenzo Librandi, Apr 23 2011

with(combinat): f := n->fibonacci(n)+(-1)^n; seq(f(n), n=0..40);

Table[Fibonacci[n]+(-1)^n,{n,0,50}] (* Vladimir Joseph Stephan Orlovsky, Jul 22 2008 *)
CoefficientList[Series[1/(1-2x^2-x^3), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 10 2013 *)
LinearRecurrence[{0,2,1}, {1,0,2}, 51] (* Ray Chandler, Sep 08 2015 *)

a(n)=fibonacci(n)+(-1)^n \\ Charles R Greathouse IV, Feb 03 2014

[fibonacci(n)+(-1)^n for n in (0..50)] # G. C. Greubel, Jul 13 2019

G.f.: 1/(1 - 2*x^2 - x^3).
a(n) = 2*a(n-2) + a(n-3).
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..n-k} (-1)^(n-k-j)binomial(j, k). Diagonal sums of A059260. - Paul Barry, Sep 23 2004
From Paul Barry, Oct 04 2004: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(k, n-2k)2^(3k-n).
a(n) = Sum_{k=0..floor(n/2)} binomial(k, n-2k)2^k(1/2)^(n-2k). (End)
From Paul Barry, Oct 25 2004: (Start)
G.f.: 1/((1+x)*(1-x-x^2)).
a(n) = Sum_{k=0..n} binomial(n-k-1, k). (End)
a(n) = |1 + (-1)^(n-1)*Fibonacci(n-1)|. - Bridget Tenner, Jun 04 2008
a(n) = A000045(n) + A033999(n). - Michel Marcus, Nov 14 2013
a(n) = Fibonacci(n+1) - a(n-1), with a(0) = 1. - Franklin T. Adams-Watters, Mar 26 2014
a(n) = b(n+1) where b(n) = b(n-1) + b(n-2) + (-1)^(n+1), b(0) = 0, b(1) = 1. See also A098600.  - Richard R. Forberg, Aug 30 2014
a(n) = b(n+2) where b(n) = Sum_{k=1..n} b(n-k)*A000931(k+1), b(0) = 1. - J. Conrad, Apr 19 2017
a(n) = Sum_{j=n+1..2*n+1} F(j) mod Sum_{j=0..n} F(j) for n > 2 and F(j)=A000045(j). - Art Baker, Jan 20 2019

A239798 Decimal expansion of the midsphere radius in a regular dodecahedron with unit edges.

Original entry on oeis.org

1, 3, 0, 9, 0, 1, 6, 9, 9, 4, 3, 7, 4, 9, 4, 7, 4, 2, 4, 1, 0, 2, 2, 9, 3, 4, 1, 7, 1, 8, 2, 8, 1, 9, 0, 5, 8, 8, 6, 0, 1, 5, 4, 5, 8, 9, 9, 0, 2, 8, 8, 1, 4, 3, 1, 0, 6, 7, 7, 2, 4, 3, 1, 1, 3, 5, 2, 6, 3, 0, 2, 3, 1, 4, 0, 9, 4, 5, 1, 2, 2, 4, 8, 5, 3, 6, 0, 3, 6, 0

Offset: 1

Stanislav Sykora, Mar 27 2014

In a regular polyhedron, the midsphere is tangent to all edges.

Apart from leading digits the same as A019863 and A019827. - R. J. Mathar, Mar 30 2014

			1.30901699437494742410229341718281905886015458990288143106772431135263...

Stanislav Sykora, Table of n, a(n) for n = 1..2000
Wikipedia, Platonic solid.
Index entries for algebraic numbers, degree 2.

Cf. A000045, A001622, A001654, A019827, A019863, A066983, A094884, A104457, A187799.

Midsphere radii in Platonic solids: A020765 (tetrahedron), A020761 (octahedron), A010503 (cube), A019863 (icosahedron).

Digits:=100: evalf((3+sqrt(5))/4); # Wesley Ivan Hurt, Mar 27 2014

RealDigits[GoldenRatio^2/2,10,105][[1]] (* Vaclav Kotesovec, Mar 27 2014 *)

PARI
```
(3+sqrt(5))/4
```

Equals phi^2/2, phi being the golden ratio (A001622).
Equals (3+sqrt(5))/4.
Equals lim_{n->oo} A000045(n)/A066983(n). - Dimitri Papadopoulos, Nov 23 2023
Equals Product_{k>=2} (1 + (-1)^k/A001654(k)). - Amiram Eldar, Dec 02 2024
Equals A094884^2 = A104457/2 = 10/A187799. - Hugo Pfoertner, Dec 02 2024

A069403 a(n) = 2Fibonacci(2n+1) - 1.

Original entry on oeis.org

1, 3, 9, 25, 67, 177, 465, 1219, 3193, 8361, 21891, 57313, 150049, 392835, 1028457, 2692537, 7049155, 18454929, 48315633, 126491971, 331160281, 866988873, 2269806339, 5942430145, 15557484097, 40730022147, 106632582345, 279167724889, 730870592323, 1913444052081

Offset: 0

R. H. Hardin, Mar 22 2002

Half the number of n X 3 binary arrays with a path of adjacent 1's and a path of adjacent 0's from top row to bottom row.

Indices of A017245 = 9*n + 7 = 7, 16, 25, 34, for submitted A153819 = 16, 34, 88,. A153819(n) = 9*a(n) + 7 = 18*F(2*n+1) -2; F(n) = Fibonacci = A000045, 2's = A007395. Other recurrence: a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3). - Paul Curtz, Jan 02 2009

Colin Barker, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 41, 56.
J. Hietarinta and C.-M. Viallet, Singularity confinement and chaos in discrete systems, Physical Review Letters 81 (1998), pp. 326-328.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000045, A084707.
Cf. 1 X n A000225, 2 X n A016269, vertical path of 1 A069361-A069395, vertical paths of 0+1 A069396-A069416, vertical path of 1 not 0 A069417-A069428, no vertical paths A069429-A069447, no horizontal or vertical paths A069448-A069452.
Equals A052995 - 1.
Bisection of A001595, A062114, A066983.

List([0..30], n-> 2*Fibonacci(2*n+1)-1); # G. C. Greubel, Jul 11 2019

[2*Fibonacci(2*n+1)-1: n in [0..30]]; // Vincenzo Librandi, Apr 18 2011

a[n_]:= a[n] = 3a[n-1] - 3a[n-3] + a[n-4]; a[0] = 1; a[1] = 3; a[2] = 9; a[3] = 25; Table[ a[n], {n, 0, 30}]
Table[2*Fibonacci[2*n+1]-1, {n,0,30}] (* G. C. Greubel, Apr 22 2018 *)
LinearRecurrence[{4,-4,1},{1,3,9},30] (* Harvey P. Dale, Sep 22 2020 *)

a(n) = 2*fibonacci(2*n+1)-1 \\ Charles R Greathouse IV, Jun 11 2015

Vec((1-x+x^2)/((1-x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 02 2016

[2*fibonacci(2*n+1)-1 for n in (0..30)] # G. C. Greubel, Jul 11 2019

a(0) = 1, a(1) = 3, a(2) = 9, a(3) = 25; a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4).
a(n) = 3*a(n-1) - a(n-2) + 1 for n>1, a(1) = 3, a(0) = 0. - Reinhard Zumkeller, May 02 2006
From R. J. Mathar, Feb 23 2009: (Start)
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: (1-x+x^2)/((1-x)*(1-3*x+x^2)). (End)
a(n) = 1 + 2*Sum_{k=0..n} Fibonacci(2*k) = 1+2*A027941(n). - Gary Detlefs, Dec 07 2010
a(n) = (2^(-n)*(-5*2^n -(3-sqrt(5))^n*(-5+sqrt(5)) +(3+sqrt(5))^n*(5+sqrt(5))))/5. - Colin Barker, Nov 02 2016

Simpler definition from Vladeta Jovovic, Mar 19 2003

A095344 Length of n-th string generated by a Kolakoski(9,1) rule starting with a(1)=1.

Original entry on oeis.org

1, 1, 9, 9, 49, 81, 281, 601, 1729, 4129, 11049, 27561, 71761, 182001, 469049, 1197049, 3073249, 7861441, 20154441, 51600201, 132217969, 338618769, 867490649, 2221965721, 5691928321, 14579791201, 37347504489, 95666669289, 245056687249, 627723364401

Offset: 1

Benoit Cloitre, Jun 03 2004

Each string is derived from the previous string using the Kolakoski(9,1) rule and the additional condition: "string begins with 1 if previous string ends with 9 and vice versa". The strings are 1 -> 9 -> 111111111 -> 919191919 -> 11111111191111111119... -> ... and each one contains 1,1,9,9,31,... elements.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,5,4).

Cf. A000002, A066983, A095342, A095343.

Cf. A123270, A090390.

a:=[1,1,9];; for n in [4..35] do a[n]:=5*a[n-2]+4*a[n-3]; od; a; # G. C. Greubel, Dec 26 2019

a095344 n = a095344_list !! (n-1)
a095344_list = tail xs where
   xs = 1 : 1 : 1 : zipWith (-) (map (* 5) $ zipWith (+) (tail xs) xs) xs
-- Reinhard Zumkeller, Aug 16 2013

R:=PowerSeriesRing(Integers(), 35); Coefficients(R!( x*(1+x+ 4*x^2)/((1+x)*(1-x-4*x^2)) )); // G. C. Greubel, Dec 26 2019

seq(simplify(2*(-1)^n -(2/I)^n*(ChebyshevU(n, I/4) -2*I*ChebyshevU(n-1, I/4)) ), n = 1..35); # G. C. Greubel, Dec 26 2019

Table[2*(-1)^n - 2^n*(Fibonacci[n+1, 1/2] - 2*Fibonacci[n, 1/2]), {n,35}] (* G. C. Greubel, Dec 26 2019 *)
LinearRecurrence[{0,5,4},{1,1,9},40] (* Harvey P. Dale, Oct 12 2022 *)

Vec(x*(1+x+4*x^2)/((1+x)*(1-x-4*x^2)) + O(x^50)) \\ Colin Barker, Apr 20 2016

vector(35, n, round( 2*(-1)^n - (2/I)^n*(polchebyshev(n, 2, I/4) -2*I*polchebyshev(n-1, 2, I/4)) )) \\ G. C. Greubel, Dec 26 2019

def A095344_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x*(1+x+4*x^2)/((1+x)*(1-x-4*x^2)) ).list()
a=A095344_list(35); a[1:] # G. C. Greubel, Dec 26 2019

a(1) = a(2) = 1; for n>1, a(n) = a(n-1) + 4*a(n-2) - 4*(-1)^n.
G.f.: x*(1 + x + 4*x^2)/((1 + x)*(1 - x - 4*x^2)). - Colin Barker, Mar 25 2012
a(n) = 5*a(n-2) + 4*a(n-3). - Colin Barker, Mar 25 2012
a(n) = 2*(-1)^n + (2^(-1-n)*(-(-7+sqrt(17))*(1+sqrt(17))^n - (1-sqrt(17))^n*(7+sqrt(17))))/sqrt(17). - Colin Barker, Apr 20 2016
a(n) = 2*(-1)^n - 2^n*(Fibonacci(n+1, 1/2) - 2*Fibonacci(n, 1/2)) = 2*(-1)^n - (2/I)^n*(ChebyshevU(n, I/4) - 2*I*ChebyshevU(n-1, I/4)). - G. C. Greubel, Dec 26 2019

A108103 Fixed point of the square of the morphism: 1->3, 2->1, 3->121, starting with 1.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1

Offset: 1

Roger L. Bagula, Jun 03 2005

nonn

Old name was: A Fibonacci like substitution for three-symbol substitution with characteristic polynomial: x^3-2*x-1.
This sequence gives a three-symbol substitution for A095345.
From Michel Dekking, Jan 06 2018: (Start) What is probably meant by this statement is that A095345 is a morphic sequence, i.e., a letter-to-letter projection of a fixed point of the morphism tau given by tau(1)=121, tau(2)=3, tau(3)=313, followed by the morphism pi given by pi(1)=1, pi(2)=1, pi(3)=3.
This deserves a proof. In fact a proof can only be given if one formulates a joint statement about the sequences v:=A095345=1113111313... and w:=A095346=3131113...., because these two sequences are defined in a loop. Let D be the so-called differentiation operator which maps a word to the lengths of its runs, as studied in [Dekking, 1981]. For example D(1113111) = 313.
The words v and w by definition satisfy D(v)=w, D(w)=v. They are in fact points of period 2 for D (cf. [Dekking,1995]).
Claim: v=A095345 equals pi(x), where x is the fixed point of tau with x(1)=1, and w=A095346 equals pi(y), where y is the fixed point of tau with y(1)=3.
Proof: This is easily shown by induction on n=2,3,..., proving that tau^(n+1)(1) and tau^(n)(3) satisfy D(pi(tau^(n+1)(1)) = pi(tau^n(3)) & D(pi(tau^n(3)) = pi(tau^n(1)).
(End)
Real Salem Roots: {{x -> -1.}, {x -> -0.618034}, {x -> 1.61803}}.
From Michel Dekking, Dec 27 2017: (Start)
Let tau be the morphism squared: tau(1)=121, tau(2)=3, tau(3)=313.
Then tau(a)=a.
Claims:
(A) a(2n-1) = 1 for n = 1,2,....
(B) a(2n) = A282162(n-1) for n = 1,2,....
Proof of (A): Obviously 2 only occurs in 121, but this implies that also 3 only occurs in 131.
Proof of (B): Let R be the 'remove 1' operator, e.g.,  R(12131) = 23.
Let psi be the square of the Fibonacci morphism on the alphabet {3,2}: psi(3)=323, psi(2)=32. One proves by induction that R(tau^k(1))3 = 2psi^(k-1)(3) and R(tau^k(2))2 = 3psi^(k-1)(2) for k=1,2,.... This implies (B): see CROSSREFS in A282162.
We give the more complicated induction step of the two:
       R(tau^(k+1)(2))2 = R(tau^k(3))2 =
       R(tau^(k-1)(3))R(tau^(k-1)(1))R(tau^(k-1)(3))2 =
       R(tau^k(2))2psi^(k-2)(3)3^(-1)R(tau^k(2))2 =
       3psi^(k-2)(2)psi^(k-2)(3)psi^(k-1)(2) = 3psi^(k-2)(32332)=
       3psi^k(2).
(End)

F. M. Dekking: "What is the long range order in the Kolakoski sequence?" in: The Mathematics of Long-Range Aperiodic Order, ed. R. V. Moody, Kluwer, Dordrecht (1997), pp. 115-125.

Michel Dekking, Table of n, a(n) for n = 1..1200
F. M. Dekking, On the structure of self-generating sequences, Seminar on Number Theory, 1980-1981 (Talence, 1980-1981), Exp. No. 31, 6 pp., Univ. Bordeaux I, Talence, 1981. Math. Rev. 83e:10075.
F. M. Dekking, What Is the Long Range Order in the Kolakoski Sequence?, Report 95-100, Technische Universiteit Delft, 1995.

Cf. A000045, A066983, A282162, A095345, A095346.

s[1] = {3}; s[2] = {1}; s[3] = {1, 2, 1}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n - 1]] a = p[12]
Nest[Flatten[# /. {1 -> {3}, 2 -> {1}, 3 -> {1, 2, 1}}] &, {1}, 10] (* Robert G. Wilson v, Nov 05 2015 *)

from math import isqrt
def A108103(n): return 1 if n&1 else 1+((k:=n>>1)+isqrt(m:=5*k**2)>>1)-(k-1+isqrt(m-10*k+5)>>1) # Chai Wah Wu, May 05 2025

1->121, 2->3, 3->313.

New name from Joerg Arndt, Jan 17 2013
New name from Robert G. Wilson v, Nov 05 2015
Name corrected by Michel Dekking, Dec 27 2017
Offset 1 from Michel Dekking, Jan 01 2020

A095342 Number of elements in n-th string generated by a Kolakoski(5,1) rule starting with a(1)=1.

Original entry on oeis.org

1, 1, 5, 5, 17, 25, 61, 109, 233, 449, 917, 1813, 3649, 7273, 14573, 29117, 58265, 116497, 233029, 466021, 932081, 1864121, 3728285, 7456525, 14913097, 29826145, 59652341, 119304629, 238609313, 477218569, 954437197, 1908874333, 3817748729

Offset: 1

Benoit Cloitre, Jun 03 2004

nonn

Each string is derived from the previous string using the Kolakoski(5,1) rule and the additional condition: "string begins with 1 if previous string ends with 5 and vice versa". The strings are 1 -> 5 -> 11111 -> 51515 -> 11111511111511111 -> ... and each one contains 1,1,5,5,17,... elements.

Equals inverse binomial transform of A025579. - Gary W. Adamson, Mar 04 2010

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,2).

Cf. A000002, A066983, A095343, A095344.

Cf. A025579 . - Gary W. Adamson, Mar 04 2010

List([1..35], n-> (2^(n+2) + (-1)^n*(5-6*n))/9); # G. C. Greubel, Dec 26 2019

[(2^(n+2) + (-1)^n*(5-6*n))/9: n in [1..35]]; // G. C. Greubel, Dec 26 2019

seq( (2^(n+2) + (-1)^n*(5-6*n))/9, n=1..35); # G. C. Greubel, Dec 26 2019

Table[(2^(n+2) + (-1)^n*(5-6*n))/9, {n,35}] (* G. C. Greubel, Dec 26 2019 *)

vector(35, n, (2^(n+2) + (-1)^n*(5-6*n))/9) \\ G. C. Greubel, Dec 26 2019

[(2^(n+2) + (-1)^n*(5-6*n))/9 for n in (1..35)] # G. C. Greubel, Dec 26 2019

a(1) = a(2) = 1, a(n) = a(n-1) + 2*a(n-2) - 2*(-1)^n.
From R. J. Mathar, Apr 01 2010: (Start)
G.f.: x*(1+x+2*x^2)/((1-2*x)*(1+x)^2).
a(n) = (2^(n+2) + (-1)^n*(5-6*n))/9. (End)
E.g.f.: (exp(2*x) - 9 + (5+6*x)*exp(-x))/9. - G. C. Greubel, Dec 26 2019

cp's OEIS Frontend

A121364 Convolution of A066983 with the double Fibonacci sequence A103609.

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A153864 Triangle read by rows, A000012 * A153860 * (A066983 * 0^(n-k)).

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A111091 Successive generations of a Kolakoski(3,1) rule starting with 1 (see A066983).

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A052952 a(n) = Fibonacci(n+2) - (1-(-1)^n)/2.

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A008346 a(n) = Fibonacci(n) + (-1)^n.

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A239798 Decimal expansion of the midsphere radius in a regular dodecahedron with unit edges.

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A069403 a(n) = 2*Fibonacci(2*n+1) - 1.

A069403 a(n) = 2Fibonacci(2n+1) - 1.